Ambrosio L., Caffarelli L., Crandall M.G., Evans L.C., Fusco N. Calculus of Variations and Nonlinear Partial Differential Equations

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90 M.G. Crandall

(b) If u satisﬁes the conditions of (a), then u is locally Lipschitz continuous

in U.

comparison with cones from above.

(d) If u satisﬁes the conditions of (a), then the quantity

(y, r):= max

{w:|w−y|≤r}



u(w) −u(y)



=max

{w:|w−y|=r}



u(w) −u(y)



(4.3)

is nonnegative and nondecreasing in r, 0 <r<dist (y, ∂U). Moreover,

if |w − y| = r and S

(y, r)=

u(w) −u(y)

, then

(y, r) ≤ S

(w, s) for 0 <s<dist (y, ∂U) − r.

(4.4)

(e) u satisﬁes the conditions of (a) if and only if for every y ∈ U

r → max

(y)

(4.5)

is convex on 0 ≤ r<dist (y, ∂U).

Proof. Assume that y ∈ U, (4.1) holds,

(y) ⊂ U, |x − y| <rand u(x)=

max

(y)

u. Then we may replace u(w)byu(x) in (4.1) to conclude that

u(x)(1 −|x − y|/r) ≤ u(y)(1 −|x − y|/r),

which implies that u(x) ≤ u(y). Since also u(x) ≥ u(y), we conclude that

u(x)=u(y). Sincethisistrueforally such that x ∈ B

(y)  U and u(x)=

max

(y)

u, it is true if B

(x) ⊂ U, u(x)=max

(x)

u and |y − x| <R/2.

Thus if u has a local maximum point, it is constant in a ball around that

point. We record this: if

(x)  U ,

u satisﬁes (4.1) and u(x)= max

(x)

u, then u is constant on B

R/2

(x) . (4.6)

This guarantees that if u assumes its maximum value at any point of a con-

nected open set, then it is constant in that set, and hence that the maximum

of u over any closed ball is attained in the boundary. There is no diﬀerence

between (4.1) and (2.16) and (a) is proved.

We turn to (b). Assume, to begin, that u ≤ 0. Then, as u(w) ≤ 0 in (4.1),

the u(w) on the right can be dropped. Thus we have, written three equivalent

ways,

(a) u(x) ≤



1 −

|x − y|



u(y);

(b) − u(y) ≤−u(x)



r −|x − y|



;

|x − y|

u(y).

(4.7)

A Visit with the ∞-Laplace Equation 91

Either of (4.7) (a) or (b) is a Harnack inequality; (b) is displayed just

because you might prefer the nonnegative function −u, which is lower-

semicontinuous and enjoys comparison with cones from below, to u. If u(x) =0,

either estimates the ratio u(y)/u(x) by quantities not depending on u. Taking

the limit inferior as y → x on the right of (4.7)(a), we ﬁnd that u is lower-

semicontinuous as well as upper-semicontinuous, so it is continuous. If also

(x)  U, we may interchange x and y in (4.7) (c) and conclude, from the

two relations, that

|u(x) − u(y)|≤−min(u(x),u(y))

|x − y|

r −|x − y|

As u is locally bounded, being continuous, we conclude that it is also lo-

cally Lipschitz continuous. If u ≤ 0 does not hold and x, y ∈ B

(z) , where

(z) ⊂U, replace u by u − max

(z)

u. We thus learn that u is Lipschitz

continuous in B

(z)ifB

(z) ⊂ U.

We turn to (c). The assumptions of (a) imply that if |x −y| = s ≤ r, then

u(x) − u(y)

≤ max

{w:|w−y|=r}



u(w) −u(y)



The monotonicity of S

(r, y)inr follows upon maximizing the left-hand side

with respect to x, |x −y| = s. The quantity S

(y, r) is nonnegative by what

was already shown - u attains its maximum over a ball on the boundary.

To prove (d), let the assumptions of (4.4) hold: r<dist (y, ∂U)and

|w − y| = r, u(w)= max

(y)

u. (4.8)

Let 0 <s<dist (y, ∂U) and for 0 ≤ t ≤ 1 put y

:= y + t(w − y). By our

assumptions,

u(y

) − u(y) ≤



u(w) −u(y)



− y| = t(u(w) − u(y));

equivalently,

u(w) −u(y)

|w − y|

≤

u(w) −u(y

)

|w − y

which implies, using the choice of w and monotonicity of S

, that

(y, r)=

u(w) −u(y)

|w − y|

≤

u(w) −u(y

)

|w − y

≤ S

,s)

for s ≥|w − y

| =(1− t)|w − y|. Letting t ↑ 1 and using the continuity of

(x, s)inx, this yields

(y, r) ≤ S

(w, s).

92 M.G. Crandall

We turn to (e). If

M(r):= max

|w−y|≤r

u(w)isconvexon0≤ r<dist (y, ∂U) (4.9)

we note that then M(0) = u(y)andso

M(s) − M (0)

≤

M(r) − M (0)

for 0 <s≤ r (4.10)

says that

u(x) ≤ u(y)+ max

{w:|w−y|≤r}



u(w) −u(y)



for |x − y|≤s, and (4.1) holds.

To prove the converse, recall that if (4.1) holds, then so does (2.16). By

the the proof of Section 2.2, u is ∞-subharmonic. By Section 2.3, u enjoys

comparison with cones from above. Therefore we have the following variant

of (2.16):

u(x) ≤



max

|w−y|≤r

u(w)



|x − y|−s

r −s



max

|w−y|≤s

u(w)



r −|x − y|

r −s

(4.11)

for 0 ≤ s ≤|x −y|≤r. The inequality is obvious for |x −y| = s, r, so it holds

as asserted. Taking the maximum of u(x)over

(y), where s ≤ τ ≤ r yields

max

|w−y|≤τ

u(w) ≤



max

|w−y|≤r

u(w)



τ − s

r − s



max

|w−y|≤s

u(w)



r −τ

r −s

(4.12)

which says that

r → max

|w−y|≤r

u(w) is convex. (4.13)

We now know that (2.16) guarantees that u is locally Lipschitz continuous.

To use this fact and the estimate implicit in (2.16) eﬃciently, we introduce

the local Lipschitz constant L(v, x) of a function v at a point x.

Deﬁnition 4.1. Let v : U → IR a n d x ∈ U. Then

L(v, x):= lim

r↓0

Lip(v, B

(x)) = inf

0<r<dist (x,∂U )

Lip(v, B

(x)).

(4.14)

Of course, L(u, x)=∞ is quite possible.

Lemma 4.2. Let v : U → IR .

(a) L(v, x) is upper-semicontinuous in x ∈ U.

(b) If v is diﬀerentiable at y ∈ U, then then L(v, y) ≥|Dv(y)|.

A Visit with the ∞-Laplace Equation 93

(d) Let the line segment [y,z] ⊂ U. Then

|v(y) − v(z)|≤



max

w∈[y,z]

L(v, w)



|y −z|.

In consequence, Lip(v, B

(y)) ≤ max

x∈B

(y)

L(v, x).

(e) Dv ∈ L

∞

(U) holds in the sense of distributions if and only if L(v, x) is

bounded on U and then

sup

x∈U

L(v, x)=|Dv|

∞

(U)

and L(v, x) = lim

r↓0

|Dv|

∞

(x))

. (4.15)

Proof. To establish (a), note that if x

→ x, then B

r−|x−x





⊂ B

(x)and

therefore, for large j,

L(u, x

) ≤ Lip(u, B

r−|x−x





) ≤ Lip(u, B

(x))

Let j →∞and then r ↓ 0 to conclude that lim sup

j→∞

L(u, x

) ≤ L(u, x).

We prove (b) assuming, as we may, that p:= Du(y) =0, choose x = y +λp

with small λ>0 to ﬁnd

Lip(v, B

λ|p|

(y)) ≥

|v(y + λp) − v(y)|

|λp|

≥

λ|p|

+o(1) asλ ↓ 0.

so, recalling Dv(y)=p, L(v, y) ≥|p|.

The example n =1,v(x)=x

sin(1/x)andy = 0 shows that, in general,

L(v, y) > |Dv(y)|.

Now assume that L(v, y)=0. Then for each ε>0 there exists δ>0such

that

|x − y| <δ =⇒|v(x) − v(y)|≤Lip(v, B

(y))|x − y|≤ε|x − y|,

which proves (c).

We turn to (d). For t ∈ [0, 1), 0 ≤ δ<1 −t and r>δ|z − y|, and

g(t)=|v(y + t(z −y)) − v(y)|,

we have

g(t + δ) − g(t)

≤

|v(y +(t + δ)(z − y)) − v(y + t(z − y))|

≤ Lip(v, B

(y + t(z −y)))|z −y| for r ≥ δ|z − y|.

Letting δ ↓ 0 yields

lim sup

δ↓0

g(t + δ) − g(t)

≤ Lip(v, B

(y + t(z −y)))|z −y|

and then, letting r ↓ 0,

94 M.G. Crandall

lim sup

δ↓0

g(t + δ) − g(t)

≤ L(v, y + t(z − y))|z − y|≤ max

w∈[y,z]

L(v, w)|z − y|.

This implies, using elementary facts about Dini derivatives (for example), that

g(1) − g(0) = |v(z) −v(y)|≤ max

w∈[y,z]

L(v, w)|z − y|,

as claimed.

The claim (e) follows easily from (d) if we take as known that Dv ∈ L

∞

(U)

and |Dv|

∞

(U)

= L if and only if [y, z] ⊂ U implies

|v(z) − v(y)|≤L|z − y|

and L is the least such constant.

Deﬁnition 4.2. If u is ∞-subharmonic in U and x ∈ U, then

(x):= lim

r↓0

(x, r) = lim

r↓0

max

{w:|w−x|=r}



u(w) −u(x)



. (4.16)

Similarly, if u is ∞-superharmonic, then

−

(x):= lim

r↓0

−

(x, r) := lim

r↓0

min

{w:|w−x|=r}



u(w) −u(x)



. (4.17)

Remark 4.1. Note the notational peculiarities. S

is used both with two ar-

guments and with a single argument; no confusion should arise from this once

noted, but the meaning changes with the usage. L has two arguments, and, in

contrast to S

, one of them is the function itself. (We did not want to write

(u, x, r) or some variant.) We will want to display the function argument

later, and the identities (4.18), (4.19) below will allow us to use L(u, x)in

place of S

(x),S

−

(x) when we need to indicate the function involved.

Lemma 4.3. Let u ∈ C(U) be ∞-subharmonic. Then for x ∈ U and

(x)  U,

L(u, x)=S

(x). (4.18)

In consequence, if u is ∞-harmonic in U and x ∈ U, then

−

(x)=−S

(x). (4.19)

Moreover, u ∈ C(U ) satisﬁes

L(u, x) ≤ max

{w:|w−x|≤r}



u(w) −u(x)



(4.20)

for

(x) ⊂ U if and only if u is ∞-subharmonic in U. Finally, at any point

y of diﬀerentiability of u,

|Du(y)| = S

(y)=L(u, y). (4.21)

A Visit with the ∞-Laplace Equation 95

Proof. First, via Lemma 4.3, we have that max

(x)

L(u, x)=Lip(u, B

(x)).

Thus

(x, r)= max

{w:|w−x|≤r}



u(w) −u(x)



≤ max

{w:|w−x|≤r}

L(u, w).

The upper-semicontinuity of L(u, x)inx (Lemma 4.2 (a)) then implies, upon

taking the limit r ↓ 0, that S

(x) ≤ L(u, x). To obtain the other inequality,

let [w, z] ⊂ U. By the local Lipschitz continuity of u, g(t):= u(w + t(z −w)) is

Lipschitz continuous in t ∈ [0, 1]. Fix t ∈ (0, 1) and observe that the deﬁnition

of S

implies, for small h>0,

g(t + h) − g(t)

u(w +(t + h)(z − w)) − u(w + t(z − w))

h|z − w|

|z − w|.

≤ S

(w + t(z −w),h|z − w|)|z − w|.

The inequality follows from the deﬁnition of S

(w+t(z −w),r). Sending h ↓ 0

we ﬁnd

lim sup

h↓0

g(t + h) − g(t)

≤ S

(w + t(z −w))|z − w|

≤



sup

y∈[w,z]

(y)



|z − w|.

Thus

u(z) − u(w)=g(1) − g(0) ≤



sup

y∈[w,z]

(y)



|z − w|.

Interchanging z and w we arrive at

|u(z) − u(w)|≤



sup

y∈[w,z]

(y)



|z − w|. (4.22)

Using also the monotonicity of S

(x, r)inr (Lemma 4.1), if δ>0 is small,

we have, via (4.22),

Lip(u, B





) ≤ sup

x∈B

)

(x) ≤ sup

x∈B

)

(x, δ).

Now send r ↓ 0andthenδ ↓ 0 to ﬁnd, via the continuity of S

(x, δ)forδ>0,

L(u, x

) ≤ S

,δ) ↓ S

)asδ ↓ 0;

this completes the proof of (4.18).

To establish (4.19), note that L(u, x)=L(−u, x)andthatS

−

for u is just

−S

for −u and invoke (4.18).

96 M.G. Crandall

If u is diﬀerentiable at y and |w −y| = r, then

u(w)=u(y)+Du(y),w− y +o(r)=⇒

max

{w:|w−y|=r}



u(w) −u(y)



= |Du(y)| +

o(r)

Now use (4.18) to conclude that (4.21) holds.

We next show that (4.20) implies (4.1), and thus that u is ∞-subharmonic.

Suppose that t → γ(t)isaC

curve in U. Using (4.20) with γ(t) in place of

x, one easily checks, using the local Lipschitz continuity of t → u(γ(t)), that,

almost everywhere,



u(γ(t))



≤ L(u, γ(t))|˙γ(t)|

≤ max

w∈B

(γ(t))



u(w) −u(γ(t))



|˙γ(t)| for r<dist (γ(t),∂U).

(4.23)

It is convenient to rewrite (4.23) as

u(γ(t))+

|˙γ(t)|

u(γ(t)) ≤



max

w∈B

(γ(t))

u(w)



|˙γ(t)|

for r<dist (γ(t),∂U).

(4.24)

If x ∈ B

(y)  U and γ(t)=y + t(x −y), then dist (γ(t),∂U) >r−t|x −

y|. Moreover, B

(y) ⊃ B

r−t|x−y|

(γ(t)) . We use this information in (4.24) to

deduce that

u(γ(t)) +

|x − y|

r −t|x −y|

u(γ(t)) ≤



max

|w−y|≤r

u(w)



|x − y|

r −t|x −y|

. (4.25)

This simple diﬀerential inequality taken with the “+” sign and integrated over

0 ≤ t ≤ 1 yields (4.1).

Note that the generality of the “±” above is superﬂuous; it corresponds

to reversing the direction of γ.

Exercise 4.1. Perform the integration of (4.25) referred to just above.

Exercise 4.2. Show that the map C(U )  u → L(u, x)(x ∈ U is ﬁxed) is

lower-semicontinuous but it is not continuous. Show, however, that the restric-

tion of this mapping to the set of ∞-subharmonic functions u is continuous

and (u, x) → L(u, x) is upper-semicontinuous.

Exercise 4.3. If u is ∞-subharmonic in U and u ≤ 0, use (4.20) to conclude

that

|Du(x)|≤−

u(x)

dist (x, ∂U)

if u is diﬀerentiable at x ∈ U.

A Visit with the ∞-Laplace Equation 97

Exercise 4.4. Show that (4.20) always holds with equality for cone functions

C(x)=a|x − z| with nonnegative slopes, so C is ∞-subharmonic on IR

and

(4.20) is sharp. Observe that (4.20) fails to hold for cones with negative slopes.

Exercise 4.5. Let u be ∞-subharmonic in U and u ≤ 0. It then follows from

(4.24) that

u(γ(t)) +

|˙γ(t)|

dist (γ(t),∂U)

u(γ(t)) ≤ 0. (4.26)

(i) If γ :[0, 1] → U , conclude that

u(γ(1)) ≤ exp



−



|˙γ(t)|

dist (γ(t),∂U)



u(γ(0)). (4.27)

(ii) Use (i) to show that if x, y ∈ B

(z) ⊂ B

(z) ⊂ U, then

|x−y|

R−r

u(x) ≤ u(y). (4.28)

This is another Harnack inequality.

(iii) Let H = {(x

,...,x

):x

> 0} be a half-space, ∆

∞

u ≥ 0andu ≤ 0in

H. Using (4.26), show that x

→ x

u(x

,...,x

) is nonincreasing and

→ u(x

,...,x

)/x

is nondecreasing on H. (Here one only needs to

check the sign of a derivative.)

(iv) Let u ≤ 0be∞-subharmonic in B

(0). Use (4.26) to show that if x ∈ IR

then

t →

u(tx)

R − t|x|

is nonincreasing on 0 ≤ t|x| <R.

In particular, if x ∈ B

(0) , then u(x)/(R −|x|) ≤ u(0)/R.

Remark 4.2. See Sections 8 and 9 concerning citations for Exercises 4.3 and

4.5. Clearly, in Exercise 4.5 we are showing how to organize various things in

the literature in an new eﬃcient way.

5 Existence and Uniqueness

We have nothing new to oﬀer regarding existence, and as far as we know

Theorem 3.1 of [8] remains the state of the art in this regard. The result

allows U to be unbounded and a boundary function b which grows at most

linearly.

Theorem 5.1. Let U be an open subset of IR

, 0 ∈ ∂U, and b ∈ C(∂U). Let

∈ IR ,A

≥ A

−

, and

−

|x| + B

−

≤ b(x) ≤ A

|x| + B

for x ∈ ∂U. (5.1)

Then there exists u ∈ C(

U) which is ∞-harmonic in U, satisﬁes u = b on ∂U

and which further satisﬁes

−

|x| + B

−

≤ u(x) ≤ A

|x| + B

for x ∈ U. (5.2)

98 M.G. Crandall

The proof consists of a rather straight-foward application of the Perron

method, using the equivalence between ∞-harmonic and comparison with

cones. In the statement, the assumption 0 ∈ ∂U is made to simplify notation

and interacts with the assumption (5.1). A translation handles the general

case, but this requires naming a point of ∂U. The Perron method runs by

deﬁning h

, h :IR

→ IR b y

(x)=sup{C(x): C(x)=a|x − z| + c, a < A

−

,c∈ IR,z ∈ ∂U, C ≤ b on ∂U},

h(x)=inf{C(x): C(x)=a|x − z| + c, a > A

,c∈ IR,z ∈ ∂U, C ≥ b on ∂U}.

and then showing that u below has the desired properties.

u(x):=sup



v(x):h

≤v ≤ h and v enjoys comparison with cones from above



(5.3)

See [8] for details, or give the proof as an exercise.

Uniqueness has always been a sore spot for the theory, in the sense that it

took a long time for Jensen [36] to give the ﬁrst, quite tricky, proof and then

another proof, still tricky, but more in line with standard viscosity solution

theory, was given by Barles and Busca [9]. A self-contained presentation of the

proof of [9], which does not require familiarity with viscosity solution theory,

is given in [8].

Here we give the skeleton of a third proof, from [28], in which unbounded

domains are treated for the ﬁrst time. This time, however, we fully reduce the

result to standard arguments from viscosity solution theory, and we do not

render our discussion self-contained in this regard.

The result, proved ﬁrst by Jensen and with a second proof by Barles and

Busca, is

Theorem 5.2. Let U be bounded, u, v ∈ C(

U),∆

∞

u ≥ 0 and ∆

∞

v ≤ 0 in U.

Then if u ≤ v on ∂U, we have u ≤ v in U.

We just sketch the new proof, as it applies to the case of a bounded domain.

The main point is an approximation result.

Proposition 5.1. Let U be a (possibly unbounded) subset of IR

and u ∈

U) be ∞-subharmonic. Let ε>0,

:= {x ∈ U : L(u, x) ≥ ε} and V

:= {x ∈ U : L(u, x) <ε}. (5.4)

Then there is function u

with the properties (a)-(f) below:

(a) u

∈ C(U) and u

= u on ∂U.

(b) u

= u on U

| =0(in the viscosity sense) on V

(d) L(u

,x) ≥ ε for x ∈ U.

(e) u

is ∞-subharmonic in U.

(f) u

≤ u and lim

ε↓0

= u.

A Visit with the ∞-Laplace Equation 99

Remark 5.1. In (c), “viscosity sense” means that at a maximum (minimum)

ˆx of u

has this property on IR

, while ε|x| does not. This is consistent with the

conventions of [31].

The assertions (a)-(c) are really a prescription of how to construct u

which is then given by a standard formula. The trickiest point is (e), which

we discuss below. Full details are available in [28]; see also Barron and Jensen

[13], Proposition 5.1, where related observations were ﬁrst made, although

with a diﬀerent set of details; one does not ﬁnd Proposition 5.1 in [13], but

there is an embedded proof of (e), diﬀerent from the one we will sketch. In

fact, these authors show that if u is ∞-harmonic, then the u

above solves

max(−∆

∞

,ε−|Du

|) = 0; this variational inequality played a fundamental

role in [36] (see Section 8). However, they do not consider subsolutions nor

note the approximation property (f) in this case.

We ﬁrst explain how Proposition 5.1 reduces Theorem 5.2 to a routine

citation of results in [31]. After that, we present our proof of the most subtle

point, (e), of Proposition 5.1; that discussion will also verify (d).

If we can show that

(x) − v(x) ≤ max

∂U

− v)forx ∈ U (5.5)

then (f) allows us to replace u

by u and the right and Theorem 5.2 follows

in the limit ε ↓ 0. Thus, using (d), we may assume, without further ado, that

L(u, x) ≥ ε for x ∈ U. (5.6)

Some version of the following, which is the step that allows us to take

advantage of (5.6), is used in all comparison proofs for ∆

∞

below, beginning

with the proof of Jensen [36]. This sort of nonlinear change of variables is

a standard tool in comparison theory of viscosity solutions. Let λ>0and

sup

2λ

and then deﬁne w by

u(x)=G(w(x)) := w(x) −

w(x)

and w(x) <

. (5.7)

That is, w(x)=(1−

√

1 − 2λu)/λ. Clearly w → u uniformly as λ ↓ 0. It

therefore suﬃces to show that

w(x) − v(x) ≤ max

∂U

(w − v)forx ∈ U (5.8)

To this end, one formally computes that

0 ≤∆

∞

u =



uDu, Du





(w)(Dw ⊗ Dw)+G



(w)D





(w)Dw, G



(w)Dw

= G



(w)G



(w)

|Dw|

+ G



(w)



wDw, Dw

(5.9)